542. 01 Matrix
Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell. The distance between two adjacent cells is 1. Example 1: Input: 0 0 0 0 1 0 0 0 0 Output: 0 0 0 0 1 0 0 0 0 Example 2: Input: 0 0 0 0 1 0 1 1 1 Output: 0 0 0 0 1 0 1 2 1 Note: The number of elements of the given matrix will not exceed 10,000. There are at least one 0 in the given matrix. The cells are adjacent in only four directions: up, down, left and right.
矩阵的bfs, 考察:
1.哪些是先入队的? 外围? 还是遍历所有符合条件的? 此处改成Integer.Max_Value,
2.入队之后, 邻居元素怎么改让其入队. 怎么跳过不符合题意的元素, visited[] 是否使用
3一般用在改矩阵的题中 tips, 可以该元素e.g. 将'o' 改成'$'(见130. Surrounded Regions)
public int[][] updateMatrix(int[][] matrix) { int m = matrix.length; int n = matrix[0].length; Queue<int[]> queue = new LinkedList<>(); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (matrix[i][j] == 0) { queue.offer(new int[]{i, j}); } else { matrix[i][j] = Integer.MAX_VALUE; } } } int[][] dirs = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}}; while (!queue.isEmpty()) { int[] cur = queue.poll(); for (int[] cell : dirs) { int r = cur[0] + cell[0]; int c = cur[1] + cell[1]; if (r < 0 || r >= m || c < 0 || c >= n || matrix[r][c] <= matrix[cur[0]][cur[1]] + 1) { continue; } matrix[r][c] = matrix[cur[0]][cur[1]] + 1; queue.offer(new int[]{r,c}); } } return matrix; }
矩阵, 常将元素构造类使用: queue.offer(new int[]{r,c}) 此处可以构造类